$$ \begin{aligned} & \alpha=\int_{\frac{1}{2}}^2 \frac{\tan ^{-1} x}{2 x^2-3 x+2} d x ........(i) \\ & \text { Let } x=\frac{1}{t} \\ & \qquad d x=-\frac{1}{t^2} d t \\ & \alpha=\int_2^{\frac{1}{2}} \frac{\tan ^{-1}\left(\frac{1}{t}\right)}{\frac{2}{t^2}-\frac{3}{t}+2}\left(\frac{-1}{t^2}\right) d t \end{aligned} $$

$$ \begin{aligned} & \alpha=\int_2^{\frac{1}{2}} \frac{\tan ^{-1}\left(\frac{1}{t}\right)}{\frac{2}{t^2}-\frac{3}{t}+2}\left(\frac{-1}{t^2}\right) d t \\ & \alpha=\int_{\frac{1}{2}}^2 \frac{\cot ^{-1} t}{2 t^2-3 t+2} d t .......(ii) \end{aligned} $$

Now by (i) + (ii)

$$ \begin{aligned} & 2 \alpha=\int_{\frac{1}{2}}^2 \frac{\frac{\pi}{2}}{2 x^2-3 x+2} d x \\ & \alpha=\frac{\pi}{8} \int_{\frac{1}{2}}^2 \frac{d x}{x^2-\frac{3 x}{2}+1} \\ & \alpha=\frac{\pi}{8} \int_{\frac{1}{2}}^2 \frac{d x}{\left(x-\frac{3}{4}\right)^2+\frac{7}{16}} \\ & \alpha=\frac{\pi}{8 \times \frac{\sqrt{7}}{4}}\left[\tan ^{-1}\left(\frac{x-\frac{3}{4}}{\frac{\sqrt{7}}{4}}\right)\right]_{\frac{1}{2}}^2 \end{aligned} $$

$\begin{aligned} & \alpha=\frac{\pi}{2 \sqrt{7}}\left[\tan ^{-1} \frac{4 x-3}{\sqrt{7}}\right]_{\frac{1}{2}}^2 \\ & \alpha=\frac{\pi}{2 \sqrt{7}}\left[\tan ^{-1} \frac{5}{\sqrt{7}}-\tan ^{-1}\left(-\frac{1}{\sqrt{7}}\right)\right] \\ & \alpha=\frac{\pi}{2 \sqrt{7}} \tan ^{-1} \frac{\left(\frac{5}{\sqrt{7}}+\frac{1}{\sqrt{7}}\right)}{1-\frac{5}{7}} \\ & \alpha=\frac{\pi}{2 \sqrt{7}} \tan ^{-1}(3 \sqrt{7}) \end{aligned}$

Now $\sqrt{7} \tan \left(\frac{2 \sqrt{7} \alpha}{\pi}\right)$

$$ \begin{aligned} & \sqrt{7} \times \tan \left(\tan ^{-1}(3 \sqrt{7})\right) \\ & \sqrt{7} \times 3 \sqrt{7} \\ & =21 \end{aligned} $$